How do you solve #8/6=(k-4)/(k-3)#?

1 Answer
Feb 22, 2017

Answer:

See the entire solution process below:

Explanation:

Multiply each side of the equation by #color(red)(6)(color(blue)(k - 3))# to eliminate the fractions while keeping the equation balanced:

#color(red)(6)(color(blue)(k - 3)) xx 8/6 = color(red)(6)(color(blue)(k - 3)) xx (k - 4)/(k - 3)#

#cancel(color(red)(6))(color(blue)(k - 3)) xx 8/color(red)(cancel(color(black)(6))) = color(red)(6)cancel((color(blue)(k - 3))) xx (k - 4)/color(blue)(cancel(color(black)(k - 3)))#

#8(color(blue)(k - 3)) = color(red)(6)(k - 4)#

Next, expand the terms within parenthesis on each side of the equation:

#(8 xx color(blue)(k)) - (8 xx color(blue)(3)) = (color(red)(6) xx k) - (color(red)(6) xx 4)#

#8k - 24 = 6k - 24#

Then, add #color(red)(24)# and subtract #color(blue)(6k)# from each side of the equation to isolate the #x# term while keeping the equation balanced:

#8k - 24 + color(red)(24) - color(blue)(6k) = 6k - 24 + color(red)(24) - color(blue)(6k)#

#8k - color(blue)(6k) - 24 + color(red)(24) = 6k - color(blue)(6k) - 24 + color(red)(24)#

#2k - 0 = 0 - 0#

#2k = 0#

Now, divide each side by #color(red)(2)# to solve for #k# while keeping the equation balanced:

#(2k)/color(red)(2) = 0/color(red)(2)#

#(color(red)(cancel(color(black)(2)))k)/cancel(color(red)(2)) = 0#

#k = 0#